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A particle that has an \(8.2 \mu \mathrm{C}\) charge moves with a velocity of magnitude \(5.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) along the \(+x\) axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge could experience has a magnitude of \(0.48 \mathrm{N}\). Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.

Short Answer

Expert verified
The magnitude is approximately 0.012 T, with the field in the \(+y\) or \(-y\) direction.

Step by step solution

01

Understanding the Problem

The problem gives us a charge of a particle, its velocity, and a situation where it experiences no magnetic force. However, the maximum possible magnetic force is provided. We need to find the magnitude and direction of the magnetic field.
02

Using the Formula for Magnetic Force

The magnetic force on a charge moving in a magnetic field is given by the equation: \[ F = qvB\sin(\theta) \]where \(q\) is the charge, \(v\) is the velocity, \(B\) is the magnetic field, and \(\theta\) is the angle between the velocity vector and magnetic field. The force is zero when \(\sin(\theta) = 0\), implying that the velocity is parallel to the magnetic field.
03

Analyzing the No Force Condition

Given that the particle experiences no magnetic force, the angle \(\theta\) must be 0 or 180 degrees. This ensures the velocity vector is parallel to the magnetic field. However, to find the maximum force, we consider the scenario when \(\sin(\theta) = 1\) (i.e., \(\theta = 90^\circ\)).
04

Calculating the Magnetic Field Magnitude

To find the magnitude of the magnetic field, use the formula for maximum force:\[ F_{\text{max}} = qvB \times 1 \]Rearranging, we find:\[ B = \frac{F_{\text{max}}}{qv} \]Substituting the known values, \( q = 8.2 \times 10^{-6} \) C, \( v = 5.0 \times 10^{5} \) m/s, and \( F_{\text{max}} = 0.48 \) N:\[ B = \frac{0.48}{8.2 \times 10^{-6} \times 5.0 \times 10^{5}} \approx 0.012 \, \text{T} \].
05

Determining the Direction of the Magnetic Field

Since the force is zero in the initial situation, the magnetic field is parallel to the velocity, along the \(+x\) axis. For maximum force, the field must be perpendicular, leading to possible directions along \(+y\) or \(-y\) axis given the \(+x\) movement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic force
In physics, the magnetic force is an important concept that describes how charged particles interact with magnetic fields. It is a fundamental force that acts on particles with an electric charge when they move through a magnetic field. This force can cause charged particles to deflect, move in circular paths, or align in specific directions.

The magnetic force on a charged particle is given by the formula:
  • \( F = qvB\sin(\theta) \)
where:
  • \(F\) is the magnetic force,
  • \(q\) is the electric charge,
  • \(v\) is the velocity of the charge,
  • \(B\) is the magnetic field strength,
  • \(\theta\) is the angle between the velocity of the charge and the direction of the magnetic field.

When \( \theta = 0^\circ \) or \(180^\circ\), the force is zero because \( \sin(\theta) = 0 \). This means the velocity of the charge is parallel to the magnetic field. For maximum force, the angle \( \theta \) should be \( 90^\circ \) where \( \sin(\theta) = 1 \). This results in the charge and magnetic field being perpendicular to each other.
Charge and velocity
The concept of charge and velocity is crucial when exploring how particles react in magnetic fields. A charge, denoted by \(q\), is a property that forces particles to experience interactions due to electric and magnetic fields. In this context, we have an example of a particle with a charge of \(8.2 \mu \text{C}\) or \(8.2 \times 10^{-6} \) C.


Velocity is the speed of the particle in a specified direction. In physics, velocity is a vector, meaning it has both a magnitude and a direction. Here, the particle is moving at a speed of \(5.0 \times 10^5 \text{ m/s}\), specifically along the \(+x\) axis.

The motion of this charged particle is essential in calculating the magnetic force it can possibly experience. The formula \( F = qvB\sin(\theta) \) suggests that only the component of velocity perpendicular to the magnetic field contributes to the force. When the direction of velocity is parallel, no force is exerted, leading to zero magnetic influence.
Vector direction in physics
Understanding vector direction in physics is key to analyzing how forces affect objects. Vectors are quantities that possess both magnitude and direction, crucial in determining the behavior of forces.

In this context of magnetic fields, the vector direction determines how the velocity of a charge relates to the magnetic field direction. The vector product, or cross product, is used to calculate forces:
  • The direction of the magnetic force is perpendicular to both the velocity of the charge and the magnetic field.
  • If the magnetic field and velocity are parallel (\(\theta = 0^\circ\) or \(180^\circ\)), the magnetic force is zero.
However, when they are perpendicular (\(\theta = 90^\circ\)), the force reaches its maximum.

In our given problem, the particle moves along the \(+x\) axis with no initial force. This indicates parallel motion with the magnetic field. To find maximum possible force directions, the field must be set perpendicularly, meaning it aligns along the \(+y\) or \(-y\) axis when considering the \(+x\) velocity direction. Such considerations are vital for predicting and calculating forces in engineering and natural phenomena.

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Most popular questions from this chapter

One component of a magnetic field has a magnitude of \(0.048 \mathrm{T}\) and points along the \(+x\) axis, while the other component has a magnitude of \(0.065 \mathrm{T}\) and points along the \(-y\) axis. A particle carrying a charge of \(+2.0 \times 10^{-5} \mathrm{C}\) is moving along the \(+z\) axis at a speed of \(4.2 \times \mathrm{x}\) \(10^{3} \mathrm{m} / \mathrm{s}\) (a) Find the magnitude of the net magnetic force that acts on the particle. (b) Determine the angle that the net force makes with respect to the \(+x\) axis.

Suppose that a uniform magnetic field is everywhere perpendicular to this page. The field points directly upward toward you. A circular path is drawn on the page. Use Ampère's law to show that there can be no net current passing through the circular surface.

A charged particle enters a uniform magnetic field and follows the circular path shown in the drawing. (a) Is the particle positively or negatively charged? Why? (b) The particle's speed is \(140 \mathrm{m} / \mathrm{s},\) the magnitude of the magnetic field is \(0.48 \mathrm{T}\), and the radius of the path is \(960 \mathrm{m}\). Determine the mass of the particle, given that its charge has a magnitude of \(8.2 \times 10^{-4} \mathrm{C}\).

Review Conceptual Example 2 as an aid in understanding this problem. A velocity selector has an electric field of magnitude \(2470 \mathrm{N} / \mathrm{C}\), directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of \(6.50 \times 10^{3} \mathrm{m} / \mathrm{s},\) enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of \(+4.00 \times 10^{-12} \mathrm{C}\) enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is \(1.90 \times 10^{-9} \mathrm{N},\) pointing directly upward. What is the speed of this particle?

A positively charged particle of mass \(7.2 \times 10^{-8} \mathrm{kg}\) is traveling due east with a speed of \(85 \mathrm{m} / \mathrm{s}\) and enters a \(0.31-\mathrm{T}\) uniform magnetic field. The particle moves through one- quarter of a circle in a time of \(2.2 \times 10^{-3} \mathrm{s},\) at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

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